Download Solutions to Exam 1 Problem 1. Suppose that A and B are sets

Solutions to Exam 1
Problem 1. Suppose that A and B are sets. Recall that
A \ B = {x ∈ A | x 6∈ B}.
Prove that
(A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B).
Solution:
Suppose that x ∈ (A \ B) ∪ (B \ A). Then x ∈ A \ B and x ∈ B \ A. If x ∈ A \ B, then
x ∈ A ∪ B since x ∈ A, and x 6∈ A ∩ B since x 6∈ B. Thus x ∈ (A ∪ B) \ (A ∩ B). We have
shown that (A \ B) ∪ (B \ A) ⊂ (A ∪ B) \ (A ∩ B).
Suppose that x ∈ (A ∪ B) \ (A ∩ B). Then x ∈ A ∪ B and x 6∈ A ∩ B. Then x ∈ A
or x ∈ B. If x ∈ A, then x 6∈ A ∩ B implies x 6∈ B, so x ∈ A \ B ⊂ (A \ B) ∪ (B \ A).
If x ∈ B, then x 6∈ A ∩ B implies x 6∈ B, so x ∈ B \ A ⊂ (A \ B) ∪ (B \ A). Thus
(A ∪ B) \ (A ∩ B) ⊂ (A \ B) ∪ (B \ A) and (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B).
Problem 2.
a) Compute gcd(19, 7) using the Euclidean Algorithm (show your work).
b) Find an expression gcd(19, 7) = m19 + n7 where m, n are integers. Use your
calculation in a) and show your work.
c) Find integers a, b ∈ Z such that
(1)
b
3
a
+
=
.
7 19
133
d) Find another solution a, b ∈ Z to equation (1).
Solution: a).
gcd(19, 7) =
=
=
=
gcd(7, 5)
gcd(5, 2)
gcd(2, 1)
1.
since 19 = 2 × 7 + 5
since 7 = 5 + 2
since 5 = 2 × 2 + 1
b).
1 =
=
=
=
5−2×2
5 − 2 × (7 − 5) = 3 × 5 − 2 × 7
3 × (19 − 2 × 7) − 2 × 7
3 × 19 − 8 × 7.
c). Multiplying both sides of (1) by 133, we obtain the equation
19a + 7b = 3.
3 = 3(3 × 19 − 8 × 7) = 19 × 9 − 7 × 24,
so a = 9, b = −24 is a solution to (1).
d). 19 × 7 − 7 × 19 = 0, so
a = 9 + 7 = 16, b = −24 − 19 = −43
is another solution to (1).
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Problem 3. In this question, only give answers to parts a) and b). Suppose that S is a
set and ϕ : Z → S is a mapping. Let n ≥ 2 be an integer. Suppose that ϕ satisfies the
following condition:
(2)
ϕ(a) = ϕ(b) whenever a, b ∈ Z are such that [a]n = [b]n .
Then Φn : Zn → S given by Φn (x) = ϕ(m) if x = [m]n ∈ Zn is a well defined mapping.
In the following problems, consider the mapping ϕ : Z → Z5 given by ϕ(m) = [m]5 .
Prove your answers by verifying (2) if your answer is yes, and give a specific counterexample
to (2) if your answer is no.
a) Is Φ10 : Z10 → Z5 , given by Φ10 (x) = [m]5 if x = [m]10 well defined?
b) Is Φ3 : Z3 → Z5 , given by Φ3 (x) = [m]5 if x = [m]3 well defined?
Solution: a). Yes. Suppose a, b ∈ Z and [a]10 = [b]10 . then 10|(a − b) so a − b = c10 for
some c ∈ Z. a − b = (2c)5 so 5|(a − b). Thus [a]5 = [b]5 , and ϕ(a) = [a]5 = [b]5 = ϕ(b).
Thus Φ10 is well defined.
b). No. [0]3 = [3]3 (since 3 − 0 = 3 × 1). However, [0]5 6= [3]5 since 3 = 3 − 0 is not
divisible by 5.
Problem 4. Suppose that n ≥ 2 is a positive integer. Suppose that x ∈ Zn . Then there
exists m ∈ Z such that x = [m]n where m ∈ Z. For t ∈ Z, define tx = [tm]n . (This
is well defined; you do not need to show this). We define x to be a generator of Zn if
Zn = {tx | t ∈ Z}. Prove that x = [m]n is a generator of Zn if and only if gcd(m, n) = 1.
Solution: Suppose that gcd(m, n) = 1. Then there exist s, t ∈ Z such that sm + tn = 1.
Suppose y ∈ Zn . Then y = [u]n for some u ∈ Z. usm + utn = u, so usm ≡ u mod n, and
thus
usx = us[m]n = [usm]n = [u]n = y.
Thus x is a generator of Zn .
Now suppose x = [m]n is a generator of Zn . Then there exists t ∈ Z such that tx =
[tm]n = [1]n , so tm ≡ 1 mod n, so there exists s ∈ Z such that tm + sn = 1. Let
d = gcd(m, n). Then d divides m and d divides n, so d divides 1 since 1 is an integral
combination of m and n. Thus d = 1, since d is a positive integer.
Problem 5. Prove that there does not exist a postive integer a such that a3 = 4.
You will not receive maximum points on this problem if you use the fundamental theorem of arithmetic in your proof.
Solution 1: Suppose that a is a positive integer such that a3 = 4. Then 2|a3 . Since 2 is
a prime and a3 = a(a2 ), 2 divides one of the factors a or a2 of a3 . If 2 divides a2 , then 2
divides the factor a of a2 . Thus 2 divides a, and 23 = 8 divides a3 = 4. Thus 4 = c8 for
some c ∈ Z+ . Thus 4 = c8 ≥ 8 which is impossible. Thus there does not exist a positive
integer a such that a3 = 4.
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Solution 2. We will show that no positive integer a is a solution to a3 = 4. 13 = 1 6= 4,
so a = 1 is not a solution. Suppose a ∈ Z+ and a ≥ 2. Then
a3 = a(a2 ) ≥ 2(a2 ) ≥ 2(2a) ≥ 23 = 8 > 4.
Thus there are no solutions a ∈ Z+ with a ≥ 2. Since there are no integers a with
1 < a < 2, there are no solutions a ∈ Z+ to a3 = 4.
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